Algorithm Design

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Chapter 5 Divide and Conquer
An Approach Using Partial Substitution
There is a somewhat weaker kind of substitution one can do, in which one
guesses the overall form of the solution without pinning down the exact values
of all the constants and other parameters at the outset.
Specifically, suppose we believe that T(n)= O(n log n), but we’re not
sure of the constant inside the O(-) notation. We can use the substitution
method even without being sure of this constant, as follows. We first write
T(n) 2, and
T(n) 2 and
q = 1 separately, since they are qualitatively different from each other--and
different from the case q = 2 as well.
The Case of q > 2 Subproblems
We begin by unro!ling (5.3) in the case q > 2, following the style we used
earlier for (5.1). We will see that the punch line ends up being quite different.
Analyzing the first few levels: We show an example of this for the case
q = 3 in Figure 5.2. At the first level of recursion, we have a single
problem of size n, which takes time at most cn plus the time spent in all
subsequent recursive calls. At the next level, we have q problems, each
of size n/2. Each of these takes time at most cn/2, for a total of at most
(q/2)cn, again plus the time in subsequent recursive calls. The next level
yields q2 problems of size n/4 each, for a total time of (q2/4)cn. Since
q > 2, we see that the total work per level is increasing as we proceed
through the recursion.
Identifying apattern: At an arbitrary levelj, we have qJ distinct instances,
each of size n/2]. Thus the total work performed at level j is qJ(cn/2]) =
(q/2)icn.
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